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I am going through the functional analysis book by John B. Conway. I have a problem

If $\{h\in\mathcal{H}:||h||\leq1\}$ is compact, show that $\text{dim}\mathcal{H}<\infty$.

Where $\mathcal{H}$ is Hilbert space. My proof is following :

Let $\mathcal{E}$ be an orthonormal basis of $\mathcal{H}$ with infinite cardinality, and let $$U:=\cup_{e\in\mathcal{E}}U_e:=\cup_{e\in\mathcal{E}}\{h\in\mathcal{H}:||h-e/2||\leq 1/200\}.$$ Then $U\subset \{h\in\mathcal{H}:||h||\leq 1\}$ because for all $h\in\mathcal{H}$, $$||h-e/2||<1/200\Rightarrow||h||=||h-e/2+e/2||\leq||h-e/2||+||e/2||<1$$ for some $e\in\mathcal{E}$. Now take open cover $\{V_e\}_{e\in\mathcal{E}}$ of $U$, where $$V_i:=\{h\in\mathcal{H}:||h-e/2||< 1/100\}.$$ Then $\{V_e\}_{e\in\mathcal{E}}$ is pairwise disjoint : let $e,e'\in\mathcal{E}$, $e\neq e'$, $h\in V_e$, and $h'\in V_{e'}$. Then $$\begin{align}||h-h'||&=||h-e/2+e/2-e'/2-(h'-e'/2)||\\&\geq\bigg|||h-e/2-(h'-e'/2)||-||e/2-e'/2||\bigg|\\&=\bigg|||h-e/2-(h'-e'/2)||-\frac{\sqrt{2}}{2}\bigg|\\&\geq\bigg|\frac{\sqrt{2}}{2}-\frac{1}{50}\bigg|>0.\end{align}$$ Therefore from $U$ is closed and $U$ is not compact, $U\subset\{h\in\mathcal{H}:||h||\leq 1\}$ is not compact.

Is my proof correct?

Jingeon An-Lacroix
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    ${h:|h|<1}$ can never be compact unless your space is ${0}$. You should distinguish between $<$ and $\leq $. – Kavi Rama Murthy Jun 01 '20 at 07:20
  • @KaviRamaMurthy Yes, you are right. I edited. – Jingeon An-Lacroix Jun 01 '20 at 07:20
  • It is hard in your question to see: (1) which subset is declared to be compact as $U$ is not the same subset than the closed unit ball (2) how you get the conclusion that this set is not compact. – mathcounterexamples.net Jun 01 '20 at 07:34
  • @mathcounterexamples.net Yes, I wrote $\mathcal{H}$ instead of ${h\in\mathcal{H}:||h||\leq 1}$. I fixed it. – Jingeon An-Lacroix Jun 01 '20 at 07:49
  • Why is $U$ closed? You have written it as a countable union of closed balls, which does not imply that it is closed. That said a much much simpler proof is to note that what you call $\mathcal E$ is an infinite and discrete set, hence it admits sequences without any limit points. Since $\mathcal E$ is a subset of the unit ball the unit ball cannot be compact. – s.harp Jun 01 '20 at 09:09
  • @s.harp $U$ is closed because : For any converging sequence ${h_n}$ and it's limit $h$ (which exists since $\mathcal{H}$ is Hilbert space), if you take $N$ large enough so that for all $n\geq N$ ${h_n}_{n\geq N}$ are in the same ball (this is possible since every ball is pairwise disjoint with positive distance $\sqrt{2}/2$), then the ball is closed hence $h$ is in the ball. And thank you for the much simpler proof! That deserves the answer I think. – Jingeon An-Lacroix Jun 01 '20 at 09:16

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Your proof basically makes use of the fact that the elements of the ONB $\mathcal E$ all have mutual distance $1$. The set $U$ is a sort of blown up version of $\mathcal E/2$ and is unnecessary: Your argument that $U$ is closed works just as well for $\mathcal E$.

A more simple and direct way to exploit this fact is to consider a sequence in $\mathcal E$ hitting no element more than once (possible iff $H$ is infinite dimensional). Since all the elements of $\mathcal E$ have mutual distance $1$ there cannot be any subsequence that is Cauchy, in particular no convergent subsequence is possible. Since $\mathcal E$ is in the unit ball you find that the unit ball cannot be compact.

s.harp
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